Is Digit Count in Range easy or hard?

Digit Count in Range is considered a Hard problem because it requires understanding digit DP or positional digit counting. The brute force idea is simple, but designing the O(log n) mathematical solution requires deeper insight into number patterns.

Digit Count in Range Python/Java solution

Implement a helper function that counts occurrences of digit d from 0 to x using positional math. Call it for high and low-1, then subtract. The same logic works across Python, Java, C++, and Go since it only involves arithmetic operations and loops.

How to solve Digit Count in Range in O(log n)?

Compute a helper function count(x, d) that counts occurrences of digit d in numbers from 0 to x. Iterate through each digit position (units, tens, hundreds) and calculate contributions using the digits to the left and right. The final answer is count(high, d) minus count(low - 1, d).

What is the best approach for Digit Count in Range?

The most efficient approach uses digit DP or positional digit counting. Instead of iterating through every number, compute how many times digit d appears from 0 to high and subtract the count from 0 to low-1. This reduces the complexity to O(log n), proportional to the number of digits.

Is Digit Count in Range asked at Google/Amazon/Meta?

Digit counting and digit DP problems frequently appear in interviews at companies like Google, Amazon, and Meta. They test mathematical reasoning, number decomposition, and dynamic programming patterns over digit positions.

What data structure is used in Digit Count in Range?

The solution mainly relies on mathematical digit decomposition and sometimes a small DP table or digit array. No heavy data structures are required; the algorithm operates on digits of the number and positional calculations.

What is the time complexity of Digit Count in Range?

The optimal digit DP or positional counting solution runs in O(log n) time because it processes each digit position of the number once. Brute force solutions require O((high - low) * log n) time since each number and its digits must be examined.

Description Hints 2 Solution Video 26:35

1067. Digit Count in Range

Hard46.4% acceptance1 companyPremium elsewhere, free here

Given a single-digit integer d and two integers low and high, return the number of times that d occurs as a digit in all integers in the inclusive range [low, high].

Example 1:

Input: d = 1, low = 1, high = 13
Output: 6
Explanation: The digit d = 1 occurs 6 times in 1, 10, 11, 12, 13.
Note that the digit d = 1 occurs twice in the number 11.

Example 2:

Input: d = 3, low = 100, high = 250
Output: 35
Explanation: The digit d = 3 occurs 35 times in 103,113,123,130,131,...,238,239,243.

Constraints:

0 <= d <= 9
1 <= low <= high <= 2 * 10⁸

Asked at 1 company

Amazon

Topics

Dynamic Programming Math

Hints open one at a time

Hint 1

Define a function f(x) to get the requested sum from 1 to x. So the answer will be f(hi) - f(lo - 1)

Hint 2

In order to solve f(x) we need to do a DP over digits approach.

26:35

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